\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\section{One}
(a)
$$ f(x)-p_{1}(f;x)=\frac{f''(\xi(x))}{2}(x-x_{0})(x-x_{1}) $$
since $f(x)=\frac{1}{x}$ and $x_{0}=1, x_{1}=2$

$$ p_{1}(f;x)= -\frac{1}{2}+\frac{3}{2} $$
so
$$ \frac{1}{x}-(-\frac{1}{2}+\frac{3}{2})=\xi^{-3}(x)(x-1)(x-2) $$
$$ \xi(x)=\sqrt[3]{2x} $$
(b)

since $\sqrt[3]{2x}$ is monotonic increasing in [1,2],
$$ max\xi(x)=\sqrt[3]{4} $$
$$ min\xi(x)=\sqrt[3]{2} $$
since $f''(\xi(x))=\frac{1}{x} $,
$$ maxf''(\xi(x))=1 $$

\section{Two}
let q be a polynomial of degree n,and q satisfies 
$$q(x_{i})=\sqrt{f_{i}} $$
then, $$ q=\sum_{i=0}^n\sqrt{f_{i}}\prod_{j=0,j\not=k}^{n}\frac{x-x_{j}}{x_{k}-x_{j}} $$
let $p=q^2$,then $p(x_{i})=f_{i}$

so,$$ p=(\sum_{i=0}^n\sqrt{f_{i}}\prod_{j=0,j\not=k}^{n}\frac{x-x_{j}}{x_{k}-x_{j}})^2 $$

\section{Three}
(a)
when n=0, $$ f[t]=e^t $$

if n=k, $$ f[t,...,t+k]=\frac{(e-1)^k}{k!}e^t $$

then, $$ f[t+1,...,t+k+1]=\frac{(e-1)^k}{k!}e^{t+1} $$

when n=k+1, 

\begin{align}
    f[t,...,t+k+1] & = \frac{f[t+1,...,t+k+1]-f[t,...,t+k]}{k+1} \\
    & = \frac{(e-1)^k}{k!}\frac{e^t(e-1)}{k+1} \\
    & = \frac{(e-1)^{k+1}}{(k+1)!}e^{t+1} 
\end{align}
so the conclusion is true.

(b)
since
$$ f([0,1,...,n])=\frac{1}{n!}f^{(n)}(\xi) $$
according to (a),we have
$$\frac{(e-1)^n}{n!}=\frac{e^{\xi}}{n!} $$
so, $$ \xi=n\ln(e-1) $$
since $$\ln(e-1)>\frac{1}{2}$$
so $\xi$ is located to the right of n/2.

\section{Four}
(a)

we have the table of the divided differences
$$
\begin{array}{cccccc} 
    0 & | & 5  &    &   &  \\
    1 & | & 3  & -2 &   &  \\   
    3 & | & 5  & 1  & 1 &  \\
    4 & | & 12 & 7  & 2 & \frac{1}{4}\\
\end{array}
$$
so,$$p_{3}(f;x)&=5-2x+x(x-1)+\frac{1}{4}x(x-1)(x-3)=\frac{1}{4}x^3-\frac{9}{4}x+5 $$

(b)
$$ p'_{3}(f;x)=\frac{3}{4}x^2-\frac{9}{4} $$
let $p'_{3}(f;x)=0$,we have $$x_{min}=\sqrt{3}$$
$$ f(x)=5-\frac{3\sqrt{3}{}}{2} $$

\section{Five}
(a)

we have the table of the divided differences
$$
\begin{array}{cccccccc}
    0 & | & 0 & & & & & \\
    1 & | & 1 & 1 & & & & \\
    1 & | & 1 & 7 & 6 & & & \\
    1 & | & 1 & 7 & 21 & 15 & & \\
    2 & | & 128 & 127 & 120 & 99 & 42 & \\
    2 & | & 128 & 448 & 321 & 201 & 102 & 30 \\
\end{array}
$$
so, $$f[0,1,1,1,2,2]=30$$
(b)

$$f^{(5)}(x)=2520x^6=30 $$
so, $$x=\frac{1}{2\sqrt{21}} $$


\section{Six}
(a)
we have the table of the divided differences
$$
\begin{array}{ccccccc}
    0 & | & 1 & & & & \\  
    1 & | & 2 & 1 & & & \\  
    1 & | & 2 & -1 & -2 & & \\  
    3 & | & 0 & -1 & 0 & \frac{2}{3} & \\  
    3 & | & 0 & 0 & \frac{1}{2} & \frac{1}{4} & -\frac{5}{36} \\  
\end{array}
$$
then, $$p(x)=1+x-2x(x-1)+\frac{2}{3}x(x-1)^2-\frac{5}{36}x(x-1)^2(x-3) $$
so, $$f(2)\approx p(2)=\frac{11}{18} $$

(b)

$$f(x)-p(x)=\frac{f^{(5)}(\xi)}{5!}x(x-1)^2(x-3)^2 $$
so, $$|f(2)-p(2)|\leq \frac{M}{60}$$

\section{Seven}

when k=1, $$\vartriangle f(x)=f(x+h)-f(x)=h f[x_{0},x_{1}] $$
if the conclusion is true for k,that is, 
$$\vartriangle^{k}f(x)=k!h^k f[x_{0},x_{1},...,x_{k}] $$
then, 
\begin{align}
    \vartriangle^{k+1}f(x)&=\vartriangle^{k}f(x+h)-\vartriangle^{k}f(x) \\
    &=k!h^k (f[x_{1},x_{1},...,x_{k+1}]-f[x_{0},x_{1},...,x_{k}]) \\
    &=k!h^k (k+1)h f[x_{0},x_{1},...,x_{k+1}] \\
    &=(k+1)!h^{k+1}f[x_{0},x_{1},...,x_{k+1}]
\end{align}
so the conclusion is true,and the proof is same for $\triangledown$.

\section{Eight}
when n=1,
\begin{align}
    \frac{\partial}{\partial{x_{0}}}f[x_{0},x_{1}]&=\frac{\partial}{\partial{x_{0}}}
    \frac{f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}} \\
    &=\frac{-f'(x_{0})(x_{1}-x_{0})+f(x_{1})-f(x_{0})}{(x_{1}-x_{0})^2} \\
    &=\frac{f[x_{0},x_{1}]-f[x_{0},x_{0}]}{x_{1}-x_{0}} \\
    &=f[x_{0},x_{0},x_{1}]
\end{align}
if the conclusion id true for n-1,that is,
\begin{equation}
     \frac{\partial f[x_{0},x_{1},...,x_{n-1}]}{\partial x_{0}}=
     f[x_{0},x_{0},...,x_{n-1}]
\end{equation}
when n,
\begin{align}
    \frac{\partial f[x_{0},x_{1},...,x_{n}]}{\partial x_{0}}&=
    \frac{\partial}{\partial x_{0}}(
    \frac{f[x_{1},...,x_{n}]-f[x_{0},...,x_{n-1}]}{x_{n}-x_{0}}) \\
    &=\frac{-\frac{\partial f[x_{0},...,x_{n-1}]}{\partial x_{0}}(x_{n}-x_{0})+
    f[x_{1},...,x_{n}]-f[x_{0},...,x_{n-1}]}{(x_{n}-x_{0})^2} \\
    &=\frac{f[x_{0},...,x_{n}]-f[x_{0},x_{0},...,x_{n-1}]}{x_{n}-x_{0}} \\
    &=f[x_{0},x_{0},...,x_{n}]
\end{align}
so the conclusion is right.

\section{Nine}
Solution:

let $p(x)=a_{0}x^n+a_{1}x^{n-1}+...+a_{n} $

let $x=\frac{(b-a)t+a+b}{2} $ $(t\in[-1,1])$

so, $p(x)=q(t)$ and $q(t)=b_{0}t^n+b_{1}t^{n-1}+...+b_{n}$

$$b_{0}=\frac{a_{0}(b-a)^n}{2^n} $$
according to Corollary 2.45,
$$max_{x\in [a,b]}|p(x)|=max_{t\in [-1,1]}|q(t)|\geq \frac{b_{0}}{2^{n-1}} $$
so,$$min \ max_{x\in [a,b]}|p(x)|=\frac{a_{0}(b-a)^n}{2^{2n-1}} $$

\section{Ten}
we know that $$||\hat{p}_{n}||_{\infty}=\frac{1}{|T_{n}(a)|} $$
for k=0,1,...,n:
$$\hat{p}_{n}(x_{k})=\frac{(-1)^k}{T_{n}(a)} , \quad x_{k}=\cos(\frac{k\pi}{n})$$
if there exists p so that $||p||_{\infty}<\frac{1}{|T_{n}(a)|}$

let $q(x)=p(x)-\hat{p}_{n}(x)$,then $q(a)=0$

and $$q(x_{k})=p(x_{k})-\frac{(-1)^k}{T_{n}(a)} $$
$$q(x_{k})q(x_{k+1})<0 $$
so,there exists $\xi_{1},...,\xi_{n}$ so that
$$ q(\xi_{1})=...=q(x_{n})=0 $$
since $q(a)=0$ and $a>1$,so $a\not=x_{k}$

so,$q(x)$ has at least n+1 zero points.

Contradictory!

So the conclusion is true.


\section{Eleven}
Solution:

(a)$b_{n,k}(t)>0$ is trivial

(b)$$\sum_{k=0}^{n}b_{n,k}(t)=\sum_{k=0}^{n}\binom{n}{k}t^k(1-t)^{n-k}=(t+1-t)^n=1 $$
(c)
\begin{align}
    \sum_{k=0}^{n}k b_{n,k}(t)&=\sum_{k=0}^{n}k\binom{n}{k}t^k(1-t)^{n-k} \\
    &=\sum_{k=0}^{n}n t \binom{n-1}{k-1}t^{k-1}(1-t)^{n-k} \\
    &=n t(t+1-t)^{n-1} \\
    &=n t
\end{align}

(d)

\begin{align}
    \sum_{k=0}^{n}k^2 b_{n,k}(t)&=\sum_{k=0}^{n}k^2\binom{n}{k}t^k(1-t)^{n-k} \\
    &=\sum_{k=0}^{n} n(n-1)\binom{n-2}{k-2}t^k(1-t)^{n-k}+
    \sum_{k=0}^{n} n\binom{n-1}{k-1}t^k(1-t)^{n-k} \\
    &=n(n-1)t^2(t+1-t)^{n-2}+n t(t+1-t)^{n-1} \\
    &=n^2t^2+n t-n t^2 \\
\end{align}

\begin{align}
    \sum_{k=0}^{n}(k-n t)^2 b_{n,k}(t)&=\sum_{k=0}^{n}(k^2+n^2t^2-2k n t)b_{n,k}(t) \\
    &=n^2t^2+n t-n t^2+n^2t^2-2n^2t^2 \\
    &=n t(1-t)
\end{align}

\end{document}
